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Probability of distributions

How to find the the probability using mean and standard deviationWhat is the probability that the “smartest” of them has IQ score above 120?probability and random sampleHow do you determine which probability distribution to use?Geometric distribution and lottery problemExceedance probability of normal and logarithmic distribution by matlabCalculating probability for a normal distributionUsing the normal distribution to find the probability of getting an exact valueThe number of Bernoulli trials required to produce exactly 1 success and at least 1 failure.Find the probability of a Normal Distribution random variable

0

$begingroup$

A particular question says the average body temperature is 98.2F with standard deviation of 0.7 with normal distribution, allowing us to find probability with z-tables.

However, an additional question asks that if 10 people were selected at random (independent of each other) and X represents the number of people who have body temperatures exceeding 98.6F. What is the probability that at least 2 people have temperatures that exceed 98.6F?

Would this distribution still be Normal, as the parameters of mean and SD still apply, or is it binomial? As there is a fixed amount of trials and success/failure as limits. But then the parameters for n and p, what would be p?

probability probability-distributions

share|cite|improve this question

asked Mar 30 at 3:20

OlivettiOlivetti

32

$endgroup$

add a comment | 

0

$begingroup$

A particular question says the average body temperature is 98.2F with standard deviation of 0.7 with normal distribution, allowing us to find probability with z-tables.

However, an additional question asks that if 10 people were selected at random (independent of each other) and X represents the number of people who have body temperatures exceeding 98.6F. What is the probability that at least 2 people have temperatures that exceed 98.6F?

Would this distribution still be Normal, as the parameters of mean and SD still apply, or is it binomial? As there is a fixed amount of trials and success/failure as limits. But then the parameters for n and p, what would be p?

probability probability-distributions

share|cite|improve this question

asked Mar 30 at 3:20

OlivettiOlivetti

32

$endgroup$

add a comment | 

$begingroup$

A particular question says the average body temperature is 98.2F with standard deviation of 0.7 with normal distribution, allowing us to find probability with z-tables.

However, an additional question asks that if 10 people were selected at random (independent of each other) and X represents the number of people who have body temperatures exceeding 98.6F. What is the probability that at least 2 people have temperatures that exceed 98.6F?

Would this distribution still be Normal, as the parameters of mean and SD still apply, or is it binomial? As there is a fixed amount of trials and success/failure as limits. But then the parameters for n and p, what would be p?

probability probability-distributions

share|cite|improve this question

asked Mar 30 at 3:20

OlivettiOlivetti

32

$endgroup$

share|cite|improve this question

asked Mar 30 at 3:20

OlivettiOlivetti

32

share|cite|improve this question

asked Mar 30 at 3:20

OlivettiOlivetti

32

asked Mar 30 at 3:20

OlivettiOlivetti

32

asked Mar 30 at 3:20

OlivettiOlivetti

32

2 Answers
2

active

oldest

votes

1

$begingroup$

Given the normal distribution, $N(98.2, 0.7^2)$, the probability of someone having temp higher than 98.6 is $hatp=1-Phi(frac98.6-98.20.7)= 0.2838546$.

Sample size is $n=10$ and the probability of at least two people having higher temp is: $P = 1 - P(X=0) - P(X=1)$, where $X$ is distributed as Binomial$(n,hatp)$. Thus, $P = 1 - (1-hatp)^10 - 10 hatp(1-hatp)^9 = 0.8238783.$

share|cite|improve this answer

edited Mar 30 at 5:05

answered Mar 30 at 4:46

dnqxtdnqxt

5525

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The OP question states that the average is $98.2$ while the threshold of interest is $98.6$, so this is an unfair coin.
  $endgroup$
  – antkam
  Mar 30 at 4:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oops, I didn't notice, thanks. Will fix presently...
  $endgroup$
  – dnqxt
  Mar 30 at 4:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I dont blame you. My mom has always told me the average is $98.6$, i.e. my mom disagrees with the premise of this question. :)
  $endgroup$
  – antkam
  Mar 30 at 5:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Moms are usually right :)
  $endgroup$
  – dnqxt
  Mar 30 at 5:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. I understand the first steps, but I thought the formula for binomial distributions was n Choose x, so how does that work?
  $endgroup$
  – Olivetti
  Apr 1 at 8:55
  

  
  
  
  

 | 
show 5 more comments

0

$begingroup$

Hint: You need to use both the Normal and the Binomial in different parts of the computation.

share|cite|improve this answer

answered Mar 30 at 3:39

ErtxiemErtxiem

671112

$endgroup$

add a comment | 

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1

$begingroup$

Given the normal distribution, $N(98.2, 0.7^2)$, the probability of someone having temp higher than 98.6 is $hatp=1-Phi(frac98.6-98.20.7)= 0.2838546$.

Sample size is $n=10$ and the probability of at least two people having higher temp is: $P = 1 - P(X=0) - P(X=1)$, where $X$ is distributed as Binomial$(n,hatp)$. Thus, $P = 1 - (1-hatp)^10 - 10 hatp(1-hatp)^9 = 0.8238783.$

share|cite|improve this answer

edited Mar 30 at 5:05

answered Mar 30 at 4:46

dnqxtdnqxt

5525

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The OP question states that the average is $98.2$ while the threshold of interest is $98.6$, so this is an unfair coin.
  $endgroup$
  – antkam
  Mar 30 at 4:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oops, I didn't notice, thanks. Will fix presently...
  $endgroup$
  – dnqxt
  Mar 30 at 4:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I dont blame you. My mom has always told me the average is $98.6$, i.e. my mom disagrees with the premise of this question. :)
  $endgroup$
  – antkam
  Mar 30 at 5:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Moms are usually right :)
  $endgroup$
  – dnqxt
  Mar 30 at 5:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. I understand the first steps, but I thought the formula for binomial distributions was n Choose x, so how does that work?
  $endgroup$
  – Olivetti
  Apr 1 at 8:55
  

  
  
  
  

 | 
show 5 more comments

1

$begingroup$

Given the normal distribution, $N(98.2, 0.7^2)$, the probability of someone having temp higher than 98.6 is $hatp=1-Phi(frac98.6-98.20.7)= 0.2838546$.

Sample size is $n=10$ and the probability of at least two people having higher temp is: $P = 1 - P(X=0) - P(X=1)$, where $X$ is distributed as Binomial$(n,hatp)$. Thus, $P = 1 - (1-hatp)^10 - 10 hatp(1-hatp)^9 = 0.8238783.$

share|cite|improve this answer

edited Mar 30 at 5:05

answered Mar 30 at 4:46

dnqxtdnqxt

5525

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The OP question states that the average is $98.2$ while the threshold of interest is $98.6$, so this is an unfair coin.
  $endgroup$
  – antkam
  Mar 30 at 4:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oops, I didn't notice, thanks. Will fix presently...
  $endgroup$
  – dnqxt
  Mar 30 at 4:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I dont blame you. My mom has always told me the average is $98.6$, i.e. my mom disagrees with the premise of this question. :)
  $endgroup$
  – antkam
  Mar 30 at 5:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Moms are usually right :)
  $endgroup$
  – dnqxt
  Mar 30 at 5:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. I understand the first steps, but I thought the formula for binomial distributions was n Choose x, so how does that work?
  $endgroup$
  – Olivetti
  Apr 1 at 8:55
  

  
  
  
  

 | 
show 5 more comments

$begingroup$

Given the normal distribution, $N(98.2, 0.7^2)$, the probability of someone having temp higher than 98.6 is $hatp=1-Phi(frac98.6-98.20.7)= 0.2838546$.

Sample size is $n=10$ and the probability of at least two people having higher temp is: $P = 1 - P(X=0) - P(X=1)$, where $X$ is distributed as Binomial$(n,hatp)$. Thus, $P = 1 - (1-hatp)^10 - 10 hatp(1-hatp)^9 = 0.8238783.$

share|cite|improve this answer

edited Mar 30 at 5:05

answered Mar 30 at 4:46

dnqxtdnqxt

5525

$endgroup$

share|cite|improve this answer

edited Mar 30 at 5:05

answered Mar 30 at 4:46

dnqxtdnqxt

5525

answered Mar 30 at 4:46

dnqxtdnqxt

5525

answered Mar 30 at 4:46

dnqxtdnqxt

5525

0

$begingroup$

Hint: You need to use both the Normal and the Binomial in different parts of the computation.

share|cite|improve this answer

answered Mar 30 at 3:39

ErtxiemErtxiem

671112

$endgroup$

add a comment | 

0

$begingroup$

Hint: You need to use both the Normal and the Binomial in different parts of the computation.

share|cite|improve this answer

answered Mar 30 at 3:39

ErtxiemErtxiem

671112

$endgroup$

add a comment | 

$begingroup$

Hint: You need to use both the Normal and the Binomial in different parts of the computation.

share|cite|improve this answer

answered Mar 30 at 3:39

ErtxiemErtxiem

671112

$endgroup$

share|cite|improve this answer

answered Mar 30 at 3:39

ErtxiemErtxiem

671112

answered Mar 30 at 3:39

ErtxiemErtxiem

671112

answered Mar 30 at 3:39

ErtxiemErtxiem

671112